Best Time to Buy and Sell Stock with Transaction Fee
2017-12-04 16:18
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Best Time to Buy and Sell Stock with Transaction Fee
A.题意
Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee.You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)
Return the maximum profit you can make.
Example 1:
Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:
- Buying at prices[0] = 1
- Selling at prices[3] = 8
- Buying at prices[4] = 4
- Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
Note:
0 < prices.length <= 50000.
0 < prices[i] < 50000.
0 <= fee < 50000.
题目的大概意思就是你要对一只股票在n天内进行操作来获得最大利润,在n天内你可以卖出或买进股票但你手里最多只能持有这一只股票,你每次卖出股票都要交一笔手续费fee。
B.分析
很显然这是一道动态规划的问题,这n天的序列你你通过不断地卖出买进这个操作我们无法贪心,这里只能采用动态规划比较来获得最大利润的值,这里我们用两个数组sell[i]表示到第i天卖股票能得到的最大利润,hold表示第i天持有股票能获得的最大利润,于是递推式为sell[i] = max(sell[i - 1],hold[i - 1] + prices[i] - fee);
hold[i] = max(hold[i - 1],sell[i - 1] - prices[i]);
最后的答案为sell
C.代码实现
class Solution { public: int maxProfit(vector<int>& prices, int fee) { int size = prices.size(); int sell[size], hold[size]; memset(sell,0,sizeof(sell)); memset(hold,0,sizeof(hold)); hold[0] = -prices[0]; for (int i = 1;i < size;i++) { sell[i] = max(sell[i - 1],hold[i - 1] + prices[i] - fee); hold[i] = max(hold[i - 1],sell[i - 1] - prices[i]); } return sell[size - 1]; } };
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